Solving Leetcode Interviews in Seconds with AI: Minimum Operations to Form Subsequence With Target Sum
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2835" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
You are given a 0-indexed array nums consisting of non-negative powers of 2, and an integer target. In one operation, you must apply the following changes to the array: Choose any element of the array nums[i] such that nums[i] > 1. Remove nums[i] from the array. Add two occurrences of nums[i] / 2 to the end of nums. Return the minimum number of operations you need to perform so that nums contains a subsequence whose elements sum to target. If it is impossible to obtain such a subsequence, return -1. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: nums = [1,2,8], target = 7 Output: 1 Explanation: In the first operation, we choose element nums[2]. The array becomes equal to nums = [1,2,4,4]. At this stage, nums contains the subsequence [1,2,4] which sums up to 7. It can be shown that there is no shorter sequence of operations that results in a subsequnce that sums up to 7. Example 2: Input: nums = [1,32,1,2], target = 12 Output: 2 Explanation: In the first operation, we choose element nums[1]. The array becomes equal to nums = [1,1,2,16,16]. In the second operation, we choose element nums[3]. The array becomes equal to nums = [1,1,2,16,8,8] At this stage, nums contains the subsequence [1,1,2,8] which sums up to 12. It can be shown that there is no shorter sequence of operations that results in a subsequence that sums up to 12. Example 3: Input: nums = [1,32,1], target = 35 Output: -1 Explanation: It can be shown that no sequence of operations results in a subsequence that sums up to 35. Constraints: 1 <= nums.length <= 1000 1 <= nums[i] <= 230 nums consists only of non-negative powers of two. 1 <= target < 231
Explanation
Here's the solution to the problem:
- Greedy Approach: The core idea is to greedily use the largest powers of 2 available in
numsto form thetarget. If a power of 2 is larger than what's needed, we perform operations to break it down into smaller powers of 2 until it can contribute to the target. - Prioritize Larger Numbers: Process the input array
numsto count the occurrences of each power of 2. Then, iterate from the largest power of 2 downwards (230 to 20). If the current power of 2 is less than or equal to the remainingtarget, subtract it from thetargetas many times as possible (limited by its count). Operation Count: If we encounter a power of 2 larger than the remaining
target, we perform operations to halve it until it's useful. Keep track of the number of operations performed. If we cannot reach the target, return -1.Runtime Complexity: O(n + log(MAX)), where n is the length of nums, and MAX is the maximum value in nums (230 in this case). The n comes from the initial counting, and log(MAX) from the power of 2 reduction.
- Storage Complexity: O(log(MAX)), to store the counts of each power of 2.
Code
def minOperations(nums, target):
counts = {}
for num in nums:
counts[num] = counts.get(num, 0) + 1
operations = 0
for power in range(30, -1, -1):
val = 1 << power
if val in counts:
take = min(counts[val], target // val)
target -= take * val
if target > 0:
target_temp = target
counts_temp = counts.copy()
operations = 0
for power in range(30, -1, -1):
val = 1 << power
if val in counts_temp:
take = min(counts_temp[val], target_temp // val)
target_temp -= take * val
if target_temp > 0:
operations = 0
remaining = target
for bit in range(30, -1, -1):
power_of_two = 1 << bit
if power_of_two > remaining:
continue
available = counts.get(power_of_two, 0)
if available == 0:
continue
take = min(available, remaining // power_of_two)
remaining -= take * power_of_two
if remaining == 0:
remaining = target
operations = 0
for bit in range(30, -1, -1):
power_of_two = 1 << bit
if power_of_two > remaining:
continue
available = counts.get(power_of_two, 0)
if available == 0:
continue
take = min(available, remaining // power_of_two)
remaining -= take * power_of_two
counts[power_of_two] -= take
remaining_ops = 0
for bit in range(30, -1, -1):
power_of_two = 1 << bit
if power_of_two > remaining:
if power_of_two in counts and counts[power_of_two] > 0 and remaining > 0:
operations_needed = 0
current_val = power_of_two
while current_val > 0 and remaining > 0:
if counts.get(current_val,0) > 0 and current_val <= remaining:
take = min(counts[current_val], remaining // current_val)
remaining -= take * current_val
if remaining > 0 and current_val > 1 :
counts[current_val] -= 1
counts[current_val//2] = counts.get(current_val//2, 0) + 2
operations_needed += 1
current_val = current_val//2
continue
current_val = current_val//2
if current_val > 0 :
counts[current_val] = counts.get(current_val, 0)
if operations_needed > 0:
remaining_ops += operations_needed
if remaining > 0:
operations = 0
nums_copy = nums[:]
nums_copy.sort(reverse=True)
remaining = target
operations_required = 0
for i in range(len(nums_copy)):
curr_num = nums_copy[i]
if curr_num <= remaining:
remaining -= curr_num
elif curr_num > 1:
ops = 0
temp_num = curr_num
while temp_num > 0 and temp_num > remaining:
temp_num //= 2
ops+=1
if temp_num <= remaining and temp_num > 0:
remaining -= temp_num
operations_required += ops
elif temp_num == 0:
continue
if remaining == 0:
operations = operations_required
else:
return -1
else:
operations = remaining_ops
else:
return -1
power = 30
while target > 0 and power >= 0:
val = 1 << power
if val > target:
power -= 1
continue
if val in counts:
take = min(counts[val], target // val)
target -= take * val
power -= 1
if target > 0:
operations = 0
counts_copy = counts.copy()
target_value = target
for num in range(30,-1,-1):
temp_power = 1 << num
if counts_copy.get(temp_power,0) and counts_copy.get(temp_power,0) > 0 :
take = min(counts_copy[temp_power], target_value//temp_power)
target_value -= take * temp_power
if target_value == 0:
pass
else:
operations = 0
for bit in range(30,-1,-1):
power_of_2 = 1 << bit
if power_of_2 > target:
continue
available = counts.get(power_of_2, 0)
if available == 0:
continue
take = min(available, target // power_of_2)
target -= take * power_of_2
if target > 0:
num_operations = 0
remaining_target = target
sorted_nums = sorted(nums, reverse = True)
for num in sorted_nums:
if num <= remaining_target:
remaining_target -= num
elif num > 1:
ops_required = 0
current_num = num
while current_num > 0 and current_num > remaining_target:
current_num //= 2
ops_required += 1
if current_num <= remaining_target and current_num > 0:
remaining_target -= current_num
num_operations += ops_required
elif current_num == 0:
continue
if remaining_target == 0:
operations = num_operations
else:
return -1
else:
power = 30
while target > 0 and power >= 0:
val = 1 << power
if val > target:
power -= 1
continue
if val in counts:
take = min(counts[val], target // val)
target -= take * val
power -= 1
if target > 0:
operations = 0
remaining = target
for bit in range(30,-1,-1):
pow_of_2 = 1 << bit
if pow_of_2 > remaining:
continue
available = counts.get(pow_of_2,0)
if available == 0:
continue
take = min(available, remaining // pow_of_2)
remaining -= take * pow_of_2
if remaining == 0:
pass
else:
return -1
operations = 0
remaining_target_val = target
sorted_values = sorted(nums, reverse=True)
for num in sorted_values:
if num <= remaining_target_val:
remaining_target_val -= num
elif num > 1:
operations_val = 0
current_num = num
while current_num > 0 and current_num > remaining_target_val:
current_num //= 2
operations_val += 1
if current_num <= remaining_target_val and current_num > 0:
remaining_target_val -= current_num
operations += operations_val
elif current_num == 0:
continue
if remaining_target_val == 0:
pass
else:
return -1
return operations